#### Answer

$x\approx1.67$ mm

#### Work Step by Step

We are given that the formula $\ln(\frac{I}{I_{0}})=-kx$ can be used to solve radiation problems (where x is the depth in millimeters, I is the intensity of radiation, $I_{0}$ is the initial intensity, and k is a constant measure dependent on the material).
In this case, we are given that the intensity of the radiation passing through a lead shield is reduced to 3% of the original intensity (which means that $\frac{I}{I_{0}}=.03$) and $k=2.1$.
$\ln(.03)=-2.1x$
Divide both sides by -2.1
$x=-\frac{\ln(.03)}{2.1}\approx1.67$ mm